Write the electronic configuration of $$Sr^{2+}$$ in absence of Aufbau Principle.

$$ 1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{5}4d^1$$

$$ 1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{3}4d^3$$

The electronic configuration of $$Rb^{+}$$ in absence of Aufbau Principle, is:

$$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{4}$$

$$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{3}$$

represented as $$\overset{-}{C}H_3$$

represented as $$\overset{+}{C}H_3$$

MCQ Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 4

Electronic configuration

1 s^{2} 2 s^{2} 2 p x^{1} 2 p y^{1} 2 p z^{1}

$1s^2 2s^2 2px^1 2py^1 2pz^1$ corresponds to :

0%
$N$
0%
$Cr$
0%
$O$
0%
$F$

Explanation

Total number of electrons

=

$=$ 7. Therefore ,atomic number of the element is 7, which is the atomic number of N.

From Be to Ra, ionization energies:

0%
increases
0%
decreases
0%
remain same
0%
none of these

Explanation

B e

$Be$ to

R a

$Ra$ ,

As we go down the group in Periodic table, atomic size increase, Nuclear hold on electron decrease for an outer electron.

Thus, Ionization energy decreases down the group.

There are two elements X and Y. 'X' is the element with the highest

I P

$IP$ value. 'Y' is the first element in a group possessing strongest reducing property. If X and Y lose 1 and 2 electrons respectively, calculate the

I P

$IP$ values of the resultant species.

0%
57.2 and 122.4 eV
0%
122.4 eV and 57.2ev
0%
57.2ev and 27. 9 ev
0%
None of these

Explanation

The element with the highest

I P

$IP$ value is helium. 'X' is helium.

I A

$IA$ group elements are the strongest reducing agents. First element of

I A

$IA$ group is lithium. 'X' loses one electron from the single electron species is equal to the energy of electron in that orbit calculated by Bohr's formula.

H e^{+} \to I P v a l u e = E_{n} = - \frac{13.6}{1} \times 2^{2}

$He^+\rightarrow IP \: value=E_n=-\dfrac{13.6}{1}\times 2^2$

= 57.2 e V

$= 57.2\:eV$

L i^{+ 2} \to I P v a l u e = E_{n} = - \frac{13.6}{1} \times 3^{2}

$Li^{+2}\rightarrow IP \: value=E_n=-\dfrac{13.6}{1}\times 3^2$

= 122.4 e V

$= 122.4eV$ .

From the following, identify least reactive element:

0%
$_{8}Y^{16}$
0%
$_{10}Y^{20}$
0%
$_{11}Y^{23}$
0%
$_{9}Y^{19}$

Explanation

Element,

_{10} Y^{20}

$_{10}Y^{20}$ has the electronic configuration as 2, 8 has completely filled valence shell. So, it is least reactive.

Which alkali metal element is the most reactive?

0%
Caesium
0%
Potassium
0%
Lithium
0%
Francium

The

100^{t h}

$100^{th}$ element is named in honor of:

0%
Einstein
0%
Bohr
0%
Fermi
0%
Curie

Explanation

The

100^{t h}

${ 100 }^{ th }$ element is 'Fermium' . It is named in honor of Scientist Fermi

Structures of nuclei of three atoms A, B, and C are given below-

A has 90 protons and 146 neutrons
B has 92 protons and 146 neutrons
C has 90 protons and 148 neutrons

Based on the above data, which of these atoms are isotopes and which are isobars?

0%
A and C are isotopes; B and C are isobars
0%
A and B are isotopes; A and C are isobars
0%
B and C are isobars; A and B are isobars
0%
A and C are isotopes; A and B are isobars

Explanation

_{90}^{236} A,_{92}^{238} B,_{90}^{238} C

$^{236}_{90}A,\quad ^{238}_{92}B,\quad ^{238}_{90}C$

A and C are isotopes because they have the same atomic number.
B and C are isobars have the same mass number.

Hence, the correct option is A.

Which of the following order is wrong?

0%
$NH_3< PH_3< AsH_3$ : $Acidic\ strength$
0%
$Li< Be<B< C$ : $IE_1$
0%
$Al_2O_3< MgO< Na_2O< K_2O$ : $Basic\ strength$
0%
$Li^+< Na^+< K^+< Cs^+$ : $Ionic \:character$

Explanation

N H_{3}, P H_{3}, A s H_{3}

$NH_3, PH_3, AsH_3$ are basic due to the presence of lone pair of electrons on the central atom. but as the size increases down the group the basicity decreases due to electron cloud diffusion over large volume/size of the central atom.

Therefore the acidic strength will increase down the group as:

N H_{3} < P H_{3} < A s H_{3}

$NH_3<PH_3<AsH_3$

B. IE of elements increases from left to right in a period due to the increase in effective nuclear charge on the element.

B e

$Be$ has high IE than

B

$B$ due to its completely filled

2 s

$2s$ orbital that provides stability to the element.

Thus the order of

I E_{1}

$IE_1$ will be:

L i < B < B e < C

$Li<B<Be<C$

C. Oxides are acidic or basic due to the formation of the acidic or basic solution with water. Within a group, the basic strength of oxides increases with the size of the element and it decreases across a period. Thus

K_{2} O

$K_2O$ is more basic than

N a_{2} O

$Na_2O$ . And

M g O

$MgO$ is more basic than

A l_{2} O_{3}

$Al_2O_3$ and less than

N a_{2} O

$Na_2O$ . In fact

A l_{2} O_{3}

$Al_2O_3$ is amphoteric oxide.

Thus order of basic strength will be:

K_{2} O > N a_{2} O > M g O > A l_{2} O_{3}

$K_2O>Na_2O>MgO>Al_2O_3$

D. Down the group the size of the elements increases due to which the polarizing power of cation decreases and its covalent character decreases and ionic character increases.

ionic character order in group 1 should be:

L i^{+} < N a^{+} < K^{+} < C s^{+}

$Li^+<Na^+<K^+<Cs^+$

Hence, the correct option is

B

$\text{B}$

Ionization potential value of Cs is equal to the electron affinity value of its corresponding __________

0%
$Cs$
0%
$Cs^+$
0%
$Cs^2+$
0%
$Cs^-$

Explanation

Ionization Potential is defined as energy required to remove electron where as E.A is defined as energy released in adding an electron.

If we remove an electron from Cs it becomes

C s^{+}

$Cs^{+}$ .If we add an electron to

C s^{+}

$Cs^{+}$ then it becomes Cs.

Hence adding electron to

C s^{+}

$Cs^{+}$ is electron affinity where other is I.P both are reaching the same point.

Hence Ionization potential value of Cs is equal to the electron affinity value of its corresponding

C s^{+}

$Cs^{+}$ .

Option B is correct.

A neutral atom of an element has a nucleus with nuclear charge

11

$11$ times and mass

23

$23$ times that of hydrogen. Write the electronic configuration of the element?

0%
$2, 1$
0%
$2, 8, 1$
0%
$2, 8$
0%
$2, 8, 8, 3$

Explanation

Since the charge of the nucleus is

11

$11$ times that of the hydrogen atom, the number of electrons is

11

$11$ and electronic configuration is

2, 8, 1

$2, 8, 1$ .

Hence, Option (B) is correct.

The numerical value of ionization energy of a unipositive ion is approximately equal to_________

0%
IP value of neutral atom.
0%
EA value of neutral atom.
0%
IP value of dipositive ion.
0%
EA value of dipositive ion.

Explanation

Ionization energy of unipositive ion means removing electron from unipositive ion.

X^{+} - e^{-} \to X^{+ +}

$X^+ - e^- \rightarrow X^{++}$

And electron affinity means electron added to dipositive ion

X^{+ +} + e^{-} \to X^{+}

$X^{++} + e^- \rightarrow X^+$

Option D is correct.

Anomalous pair among the following is:

0%
boron - silicon
0%
aluminium - nickel
0%
beryllium - indium
0%
cobalt - nickel

Explanation

Correct option: $D$

Explanation:

Introduction to Mendeleev's Periodic table:

The pair of the element in the periodic table where the element occurring before has more atomic mass than the later occurring one. It is also considered to be a drawback of Mendeleev’s periodic table.

Reasons of anomalous:

The size of an atom is less
Electronegativity is very high as a result ionization enthalpy increases.
The charge to radius ratio is large.

The below image shows the anomalous pairs of elements in Mendeleev's periodic table;

Therefore, $Co-Ni$ is an anomalous pair

An atom of an element has one electron in the valence shell. It can be represented as :

0%
$_{6}A^{12}$
0%
$_{19}A^{39}$
0%
$_{13}A^{27}$
0%
$_{12}A^{24}$

Explanation

The atomic no. of

A_{6}^{12}

$A_{6}^{12}$ is 6. Its electronic configuration is

1 s^{2} 2 s^{2} 2 p^{2}

$1s^2 2s^2 2p^2$ . It has 4 electrons in valence shell.

The atomic no. of

A_{19}^{39}

$A_{19}^{39}$ is 19. Its electronic configuration is

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}

$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$ . It has 1 electron in valance shell.

The atomic no. of

A_{13}^{27}

$A_{13}^{27}$ is 13. Its electronic configuration is

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}

$1s^2 2s^2 2p^6 3s^2 3p^1$ . It has 3 electrons in valence shell.

The atomic no. of

A_{12}^{24}

$A_{12}^{24}$ is 12. Its electronic configuration is

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}

$1s^2 2s^2 2p^6 3s^2$ . It has 2 electrons in valence shell.

Therefore, the correct answer is option B.

The atomic number of Uut is:

0%
113
0%
114
0%
108
0%
115

Explanation

U n - 1; T r i - 3

$Un-1; Tri-3$ . The symbols for these are

U, T,

$U, T,$ respectively. Hence for

U u t

$Uut$ it becomes 113. The symbols above mentioned are capital only if they are the first letter otherwise symbols would be small alphabets.

Hence, the correct option is

A

$\text{A}$

Which atoms should have smaller value of

{(I E)}_{1}

${ \left( IE \right) }_{ 1 }$ ?

(a) Oxygen or sulphur
(b) Nitrogen or phosphorus

0%
(a) $S$ (b) $P$
0%
(a) $O$ (b) $P$
0%
(a) $S$ (b) $N$
0%
$None of these$

Explanation

i) In O and S, O has higher

I E_{1}

$IE_1$ because ionization enthalpy decreases as we descend in a group due to decrease of hold by nucleus on the outermost electron.
ii) In N and P, N has higher

I E_{1}

$IE_1$ because ionization enthalpy decreases as we descend in a group due to decrease of hold by nucleus on the outermost electron.

A sudden jump between the values of second and third ionization energies of an atom would be associated with the electronic configuration :

0%
${ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 6 }{ }^{ }{ }^{ }$
0%
${ 1s }^{ 2 },{ 2s }^{ 2 }{ 2p }^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 1 }$
0%
${ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 2 }$
0%
${ 1s }^{ 2 },{ 2s }^{ 2 }{ 2p }^{ 6 },{ 3s }^{ 2 }$

Explanation

Magnesium

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2}

${1s}^{2}{2s}^{2}{2p}^{6}{3s}^{2}$ is in group 2 of the Periodic Table and has first ionisation energy 737.7kJ/mol, second ionisation energy 1450.7 kJ/mol and third ionisation energy 7732.7 kJ/mol. Here the big jump occurs after the second ionisation energy. It means that there are 2 electrons which are relatively easy to remove the

{3 s}^{2}

${3s}^{2}$ electrons, while the third one is much more difficult because it comes from an inner level - closer to the nucleus and with less screening.

Which of the following pairs show reverse properties on moving along a period from left to right and from top to bottom in a group?

0%
Atomic radius and electron gain enthalpy (negative value)
0%
Nuclear charge and ionisation enthalpy
0%
Ionisation enthalpy and electron gain enthalpy (negative value)
0%
None of the above

The nuclear charge actually experienced by an electron is termed as the effective nuclear charge. The effective nuclear charge Z* actually depends on the type of shell and orbital in which electron is actually present. The relative extent to which the various orbitals penetrate the electron clouds of other orbitals is

s > p > d > f

$s>p>d>f$ (for the same value of

n

$n$ )
The phenomenon in which penultimate shell electrons act as screen or shield in between nucleus and valence shell electrons and thereby reducing nuclear charge is known as shielding effect. The penultimate shell electrons repel the valence shell electron to keep them loosely held with the nucleus. It is thus evident that more is the shielding effect, lesser is the effective nuclear charge and lesser is the ionization energy.

Ionization energy is not influenced by :

0%
Size of atom
0%
Effective nuclear charge
0%
Electrons present in inner shell
0%
Change in entropy

Which is incorrect configuration for s-block elements?

0%
$[Ar]3d^{10}\, 4s^{2}$
0%
$[Ar]3d^{10}\, 4s^{1}$
0%
Both A and B
0%
None of these

Explanation

[A r] 3 d^{10} 4 s^{2}

$[Ar]\,3d^{10}\,4s^{2}$ - Last filling electron is in

3 d

$3d$ .

[A r] 3 d^{10} 4 s^{1}

$[Ar]\,3d^{10}\,4s^{1}$ - Last filling electron is in

3 d

$3d$ .
Thus, both are d-block element.

[A r] 4 s^{1 - 2}

$[Ar]\,4s^{1-2}$ is a configuration of s-block.

Which is the correct order of ionization energies?

0%
$F^{\displaystyle-}>F>Cl^{\displaystyle-}>Cl$
0%
$F>Cl>F^{\displaystyle-}>Cl^{\displaystyle-}$
0%
$F^{\displaystyle-}>Cl^{\displaystyle-}>Cl>F$
0%
$F^{\displaystyle-}>Cl^{\displaystyle-}>F>Cl$

Explanation

Hint: The ionization energy of anion is less than the neutral atom.

Correct Answer: Option

B

$B$

Explanation:

Since it is easier to take out an electron from a negatively charged ion than from a neutral atom, the ionization energies for anions will be lesser than the corresponding neutral atoms. So the order of ionisation energy is $F>F^-$ and $Cl>Cl^-$ .
Since the ionization energy decreases on going down the group, the ionization energy order is $F>Cl$ .
So final order will be $F>Cl>F^->Cl^-$

Atomic weight of which set of elements were corrected by Mendeleev?

0%
$Li, Na, K$
0%
$U, Be, In$
0%
$Tc, Sc, Al$
0%
None of the these

Explanation

Mendeleev corrected atomic weights of Uranium, Beryllium, and Indium (

U, B e, I n

$U, Be, In$ ) based on their positions in the periodic table. Along with that he also corrected the atomic weights of Gold (

A u

$Au$ ) and Platinum (

P t

$Pt$ ).

Options

B

$B$ is correct.

First ionization energy is lowest with :

0%
lead
0%
carbon
0%
silicon
0%
tin

The term atomic radius refers to?

0%
Charge of the atom
0%
Size of the atom
0%
Distance between the centre of nucleus and the first shell
0%
Size of the outermost orbital

Refer to the Periodic table and identify a highly reactive metal in the sixth period.

0%
$Cs$
0%
$W$
0%
$Os$
0%
$Ir$

Explanation

As we move from left to right along a period, metal activity decreases. So here "

C s

$Cs$ " (alkali metal) is a highly reactive metal in the sixth period.

All others are d-block elements so they are less reactive than

C s

$Cs$ .

Differentiating electron in

K (19)

$K(19)$ goes into ns and this is s-block element. If Aufbau rule is not followed,

K (19)

$K(19)$ will be placed in:

0%
s-block
0%
p-block
0%
d-block
0%
f-block

Explanation

K (19)

$K(19)$ in the absence of Aufbau rule will have EC as:

1 s^{2}

$1s^{2}$ ,

2 s^{2} 2 p^{6}

$2s^{2}2p^{6}$ ,

3 s^{2} 3 p^{6} 3 d^{1}

$3s^{2}3p^{6}3d^{1}$ and thus differentiating electron enters into

3 d

$3d$ d-block element.

Write the electronic configuration of

S r^{2 +}

$Sr^{2+}$ in absence of Aufbau Principle.

0%
$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{5}4d^1$
0%
$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{3}4d^3$
0%
$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{6}$
0%
None of these

Explanation

According to the Aufbau principle, electrons are filled in the increasing order of

(n + l)

$(n+l)$ value.
In the absence of Aufbau principle it wont follow the order of increasing

(n + l)

$(n+l)$ value for filling up of electrons
So the electronic configuration of

S r^{2 +}

$Sr^{2+}$ (36 electrons) would be

1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{10}, 4 s^{2} 4 p^{6}

$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{6}$ .
Hence, option

C

$C$ is correct.

Use the following system of naming elements in which first alphabets of the digits are written collectively,

0	1	2	3	4	5	6	7	8	9
nil	uni	bi	tri	quad	pent	hex	sept	oct	enn

Write three-letter symbol for the elements with atomic number

106

$106$ .

0%
uh
0%
unh
0%
uunhe
0%
unhex

Explanation

Digit	1	0	6
Symbol	Uni	nil	hex
First alphabate	u	n	h

The three-letter symbol for the elements with the atomic number

106

$106$ is unh.

The element

119

$119$ has not been discovered. What would be the symbol for this element?

0%
$Uue$
0%
$UUe$
0%
$UUE$
0%
None of these

Explanation

Ununennium is also known as eka-francium or simply element

119

$119$ .

It is a hypothetical chemical element with the atomic number

119

$119$ . Its symbol is

U u e

$Uue$ .

So, the correct answer is (A)

The single-bonded metallic radius of

N a

$Na$ is 157 pm. Give a reasonable estimation of the radii, assuming that the increment between radii of different magnitudes is 60 pm. the anionic radius of

N a

$Na$ is:

0%
$217$ pm
0%
$227$ pm
0%
$237$ pm
0%
$207$ pm

Explanation

The metallic radius is always smaller than the anionic radius.
The metallic radius is

157

$157$ pm and the increment between radii of different magnitudes is

60

$60$ pm.
Hence, the anionic radius of

N a

$Na$ is

157 p m + 60 p m = 217 p m

$157\: pm+60\: pm=217 \: pm$

The electronic configuration of

R b^{+}

$Rb^{+}$ in absence of Aufbau Principle, is:

0%
$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{6}$
0%
$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{4}$
0%
$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{3}$
0%
None of these

Explanation

According to the Aufbau principle, electrons are filled in the increasing order of (n+l) value.
In the absence of Aufbau principle it wont follow the order of increasing (n+l) value for filling up of electrons .
So, the electronic configuration of

R b^{+}

$Rb^{+}$ (36 electrons) would be-

1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{10}, 4 s^{2} 4 p^{6}

$1s^{2},\, 2s^{2}2p^{6},\, 3s^{2}3p^{6}3d^{10},\, 4s^{2}4p^{6}$ .
Hence option (A) is correct.

The factors that influence the ionisation energies are :

0%
the size of the atom
0%
the charge on the nucleus
0%
how effectively the inner electron shell screen the nuclear charge
0%
all of the above

Explanation

Factors affecting ionisation potential :

Atomic size : Larger the atomic size, smaller is the ionisation potential. It is due to that the size of atom increases the outermost electrons e farther away from the nucleus and nucleus loses the attraction on that electrons and hence can be easily removed.

Effective nuclear charge

(Z_{e f f})

$( Z_{eff} )$ : Ionisation potential increases with the increase in nuclear charge between outermost electrons and nucleus.

Screening effect : Higher is the screening effect on the outer most electrons causes less attraction from the nucleus and can be easily removed, which is leading to the lower value of ionisation potential. Penetration power of sub shells and Stability of half filled and fully filled orbitals.

Which pair is different from the others?

0%
$Li-Mg$
0%
$Na-K$
0%
$Ca-Mg$
0%
$B-Al$

Explanation

L i - M g

$Li-Mg$ have a diagonal relationship between them i.e. they show various similar properties.

Whereas

N a - K

$Na-K$ belongs to Group-1 ,

C a - M g

$Ca-Mg$ belongs to Group-2 while

B - A l

$B-Al$ belongs to Group-13

Hence the correct option is 'A'

The alkalide ion is:

0%
diamagnetic
0%
represented as $\overset{-}{C}H_3$
0%
paramagnetic
0%
represented as $\overset{+}{C}H_3$

Explanation

The alkalide ions are anions of alkali metals. Known alkalide ions are $Na^{-}$ , $K^{-}$ , $Rb^{-}$ , and $Cs^{-}.$

Taking example of $Na^{-}$ :

$Na(1s^{2}2s^{2}2p^{6}4s^{1})$ $\leftarrow$ Paramagnetic due to one unpaired electron

$Na^-(1s^{2}2s^{2}2p^{6}3s^{2})$ $\leftarrow$ diamagnetic due to paired electrons.

Recently (in 2003) element with atomic number

110

$110$ has been named by IUPAC as:

0%
$Hs$
0%
$Mt$
0%
$Ds$
0%
$Sg$

Explanation

Element

110

$110$ being named darmstadtium with symbol

D s

$Ds$ by IUPAC.

Recently (in Aug 2003) two new elements have been discovered with atomic number:

0%
$113, 114$
0%
$114, 115$
0%
$115, 116$
0%
$113, 115$

Explanation

Two superheavy elements, elements

113

$113$ and

115

$115$ , were recently synthesized through a collaborative effort between scientists from the Physical and Life Sciences Directorate at the Lawrence Livermore National Laboratory and researchers from the Joint Institute for Nuclear Research at the Flerov Laboratory for Nuclear Reactions in Dubna, Russia.

Recently discovered element with atomic number

115

$115$ is:

0%
$Uun$
0%
$Uub$
0%
$Uup$
0%
$Uus$

Arrange Be, Ca, Ba, Ra in increasing order of ionisation energy:

0%
Be < Ra < Ca < Ba
0%
Ba < Ca < Ra < Be
0%
Ra < Ba < Ca < Be
0%
Ba < Ra < Ca < Be

A hydrogen like ion is an ion containing only one electron. The energies of the electron in a hydrogen like ion are given by

E_{n} = - (2.18 \times 10^{- 18}) Z^{2} (\frac{1}{n^{2}})

$E_n = - (2.18 \times 10^{-18}) Z^2 \displaystyle \left ( \frac{1}{n^2} \right )$ joules, where, n is the principal quantum number and Z is the atomic number of the element. Then the ionisation energy (in kJ mol

^{- 1}

$^{-1}$ ) of the He

^{+}

$^+$ ion is:

0%
$2.752 \times 10^3\ kJ mol^{-1}$
0%
$5.2494 \times 10^3\ kJ mol^{-1}$
0%
$4.5214 \times 10^3\ kJ mol^{-1}$
0%
$7.482 \times 10^3\ kJ mol^{-1}$

Explanation

(I E)_{H e^{+}} = (- E_{n})

$(IE)_{He^+} = (-E_n)$

= \frac{2.18 \times 10^{- 18} Z^{2}}{n^{2}} J a t o m^{- 1}

$= \displaystyle \frac{2.18 \times 10^{-18} Z^2}{n^2} J atom^{-1}$

= \frac{2.18 \times 10^{- 18} \times (2)^{2}}{1} J a t o m^{- 1}

$= \displaystyle \frac{2.18 \times 10^{-18} \times (2)^2}{1} J\ atom^{-1}$ for

H e^{+} (n = 1, Z = 2)

$He^+ (n = 1, Z = 2)$

= 2.18 \times 10^{- 18} \times 4 \times 6.02 \times 10^{23} J m o l^{- 1}

$= 2.18 \times 10^{-18} \times 4 \times 6.02 \times 10^{23} J\ mol^{-1}$

= 5.2494 \times 10^{5} J m o l^{- 1}

$= 5.2494 \times 10^5 J mol^{-1}$

= 5.2494 \times 10^{3} k J m o l^{- 1}

$= 5.2494 \times 10^3 kJ mol^{-1}$

Hence, the correct option is

B

$\text{B}$

Identify the elements in the Periodic Table:

A: Which shows diagonal relationship to $Be$

B: Which shows diagonal relationship to $Li$

0%
$A : Al , B : Mg$
0%
$A : Si , B : Mg$
0%
$A : Mg , B : Al$
0%
$A : Al , B : Si$

Explanation

In addition to horizontal and vertical trends, there is a diagonal relationship between elements such as

L i

$Li$ and

M g

$Mg$ ,

B e

$Be$ and

A l, B

$Al, B$ and

S i

$Si$ , that have an adjacent upper left/lower right relative location in the periodic table.
These pairs of elements have similar size and electronegativity, resulting in similar properties.

IE of an element does not depend on :

0%
its nuclear charge
0%
the shielding effect
0%
electron neutrality
0%
penetration effect

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

L i

$Li$ and

M g

$Mg$ are diagonal to each other in the Periodic table.

L i

$Li$ and

M g

$Mg$ shows the diagonal relationship.

It is due to the similarity in ionic sizes and charge/radius ratio of the elements.

Hence, both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Hence, the correct option is

B

$\text{B}$

The element whose electronic conguration is

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}

$1s^22s^22p^63s^2$ is :

0%
non-metal
0%
noble gas
0%
metalloid
0%
metal

Explanation

The element with electronic configuration

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}

$1s^22s^22p^63s^2$ is magnesium.

As in outer most shell, it has 2 electrons so it has a tendency to lose its outermost electron.

Hence, it is metal.

State the electronic configuration for Nitrogen [p=7, n=7].

0%
2, 3
0%
2, 4
0%
2, 6
0%
2, 5

State the electronic configuration for Helium [

p = 2, n = 2

$p=2,\ n=2$ ].

0%
$1$
0%
$3, 2$
0%
$2$
0%
$2, 4$

State the electronic configuration for Potassium [

p = 19, n = 20

$p=19,\ n=20$ ].

0%
$2, 8, 8, 1$
0%
$2, 8, 8, 2$
0%
$2, 8, 8, 3$
0%
$2, 8, 8, 4$

State the electronic configuration for Magnesium [p=

12

$12$ , n=

12

$12$ ].

0%
$2, 8, 0$
0%
$2, 8, 1$
0%
$2, 8, 2$
0%
$2, 8, 3$

'Electronic configuration' represents :

0%
arrangement of protons
0%
arrangement of neutrons
0%
arrangement of electrons
0%
arrangement of electrons, protons and neutrons

Match the statements in List I with the correct answers in list II

List I	List II
The basic unit of matter	A: Proton
The sub-atomic particle with a positive charge	B: Neon
An atom having stable electronic configuration	C: Oxygen
A molecule formed by sharing of electron [covalency]	D: Sodium
A metallic atom having unstable electronic configuration	E: Atom

0%
1-A, 2-B, 3-C, 4-D, 5-E
0%
1-C, 2-B, 3-A, 4-D, 5-E
0%
1-A, 2-C, 3-B, 4-E, 5-D
0%
1-A, 2-B, 3-D, 4-E, 5-C
0%
1-E, 2-A, 3-B, 4-C, 5-D

Explanation

Atoms are the basic building blocks of matter that make up everyday objects.
Proton has a positive charge of +1.
An atom having stable electronic configuration is neon as its octet is complete.
A molecule formed by sharing of electron is oxygen as it is formed by sharing of two electrons by each oxygen atom.
A metallic atom having unstable electronic configuration is sodium as its outer shell is incomplete

1 s^{2} 2 s^{2} p^{6} 3 s^{1}

$1s^2 2s^2p^6 3s^1$

State the electronic configuration for Neon [p=10, n=10].

0%
2, 6
0%
2, 7
0%
2, 9
0%
2, 8

What is the electronic configuration of calcium [

p = 20, n = 20

$p=20,\ n=20$ ]?

0%
$2, 8, 8, 1$
0%
$2, 8, 8, 3$
0%
$2, 8, 8, 2$
0%
$2, 8, 10$

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 4 - MCQExams.com

0:0:3

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 4 - MCQExams.com

0:0:3

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.